Question

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally​ distributed, with a mean of 6.0 ounces and a standard deviation of 0.30 ounce. Suppose that you select a sample of 36 oranges. The probability is 77​% that the sample mean amount of juice will be greater than what​ value? (round to two decimal places as needed.)

Answer 1

Solution:
Given in the question
The amount of juice squeezed from these oranges is approximately normally​ distributed with
Mean () = 6
Standard deviation() = 0.30
Number of sample = 36
Here we need to calculate the value that The probability is 77​% that the sample means the amount of juice will be greater than?
so p-value = 0.23
Z-score from Z-table is -0.74
So Juice amount can be calculated as
X = + Z-score * = 6 - 0.74 * 0.3 = 6 - 0.222 = 5.78
The probability is 77​% that the sample means the amount of juice will be greater than 5.78.